Given a n-ary tree and a number x, find and return the number of nodes which are greater than x.
Example:
In the given tree, x = 7
Number of nodes greater than x are 4.
Approach: The idea is maintain a count variable initialize to 0. Traverse the tree and compare root data with x. If root data is greater than x, increment the count variable and recursively call for all its children.
Below is the implementation of idea.
C++
// C++ program to find number of nodes // greater than x #include <bits/stdc++.h> using namespace std;
// Structure of a node of n-ary tree struct Node {
int key;
vector<Node*> child;
}; // Utility function to create // a new tree node Node* newNode( int key)
{ Node* temp = new Node;
temp->key = key;
return temp;
} // Function to find number of nodes // greater than x int nodesGreaterThanX(Node* root, int x)
{ if (root == NULL)
return 0;
int count = 0;
// if current root is greater
// than x increment count
if (root->key > x)
count++;
// Number of children of root
int numChildren = root->child.size();
// recursively calling for every child
for ( int i = 0; i < numChildren; i++) {
Node* child = root->child[i];
count += nodesGreaterThanX(child, x);
}
// return the count
return count;
} // Driver program int main()
{ /* Let us create below tree
* 5 * / | \ * 1 2 3 * / / \ \ * 15 4 5 6 */ Node* root = newNode(5);
(root->child).push_back(newNode(1));
(root->child).push_back(newNode(2));
(root->child).push_back(newNode(3));
(root->child[0]->child).push_back(newNode(15));
(root->child[1]->child).push_back(newNode(4));
(root->child[1]->child).push_back(newNode(5));
(root->child[2]->child).push_back(newNode(6));
int x = 5;
cout << "Number of nodes greater than "
<< x << " are " ;
cout << nodesGreaterThanX(root, x)
<< endl;
return 0;
} |
Java
// Java program to find number of nodes // greater than x import java.util.*;
// Class representing a Node of an N-ary tree class Node{
int key;
ArrayList<Node> child;
// Constructor to create a Node
Node( int val)
{
key = val;
child = new ArrayList<>();
}
} class GFG{
// Recursive function to find number // of nodes greater than x public static int nodesGreaterThanX(Node root, int x)
{ if (root == null )
return 0 ;
int count = 0 ;
// If current root is greater
// than x increment count
if (root.key > x)
count++;
// Recursively calling for every
// child of current root
for (Node child : root.child)
{
count += nodesGreaterThanX(child, x);
}
// Return the count
return count;
} // Driver code public static void main(String[] args)
{ /* Let us create below tree
5
/ | \
1 2 3
/ / \ \
15 4 5 6
*/
Node root = new Node( 5 );
root.child.add( new Node( 1 ));
root.child.add( new Node( 2 ));
root.child.add( new Node( 3 ));
root.child.get( 0 ).child.add( new Node( 15 ));
root.child.get( 1 ).child.add( new Node( 4 ));
root.child.get( 1 ).child.add( new Node( 5 ));
root.child.get( 2 ).child.add( new Node( 6 ));
int x = 5 ;
System.out.print( "Number of nodes greater than " +
x + " are " );
System.out.println(nodesGreaterThanX(root, x));
} } // This code is contributed by jrishabh99 |
Python3
# Python3 program to find number of nodes # greater than x # Structure of a node of n-ary tree class Node:
def __init__( self , data):
self .key = data
self .child = []
# Function to find number of nodes # greater than x def nodesGreaterThanX(root: Node, x: int ) - > int :
if root is None :
return 0
count = 0
# if current root is greater
# than x increment count
if root.key > x:
count + = 1
# Number of children of root
numChildren = len (root.child)
# recursively calling for every child
for i in range (numChildren):
child = root.child[i]
count + = nodesGreaterThanX(child, x)
# return the count
return count
# Driver Code if __name__ = = "__main__" :
ans = 0
k = 25
# Let us create below tree
# 5
# / | \
# 1 2 3
# / / \ \
# 15 4 5 6
root = Node( 5 )
(root.child).append(Node( 1 ))
(root.child).append(Node( 2 ))
(root.child).append(Node( 3 ))
(root.child[ 0 ].child).append(Node( 15 ))
(root.child[ 1 ].child).append(Node( 4 ))
(root.child[ 1 ].child).append(Node( 5 ))
(root.child[ 2 ].child).append(Node( 6 ))
x = 5
print ( "Number of nodes greater than % d are % d" %
(x, nodesGreaterThanX(root, x)))
# This code is contributed by # sanjeev2552 |
C#
// C# program to find number of nodes // greater than x using System;
using System.Collections.Generic;
// Class representing a Node of an N-ary tree public class Node
{ public int key;
public List<Node> child;
// Constructor to create a Node
public Node( int val)
{
key = val;
child = new List<Node>();
}
} class GFG{
// Recursive function to find number // of nodes greater than x public static int nodesGreaterThanX(Node root, int x)
{ if (root == null )
return 0;
int count = 0;
// If current root is greater
// than x increment count
if (root.key > x)
count++;
// Recursively calling for every
// child of current root
foreach (Node child in root.child)
{
count += nodesGreaterThanX(child, x);
}
// Return the count
return count;
} // Driver code public static void Main(String[] args)
{ /* Let us create below tree
5
/ | \
1 2 3
/ / \ \
15 4 5 6
*/
Node root = new Node(5);
root.child.Add( new Node(1));
root.child.Add( new Node(2));
root.child.Add( new Node(3));
root.child[0].child.Add( new Node(15));
root.child[1].child.Add( new Node(4));
root.child[1].child.Add( new Node(5));
root.child[2].child.Add( new Node(6));
int x = 5;
Console.Write( "Number of nodes greater than " +
x + " are " );
Console.WriteLine(nodesGreaterThanX(root, x));
} } // This code is contributed by Amit Katiyar |
Javascript
// JavaScript program to find number of nodes // greater than x // Structure of a node of n-ary tree class Node { constructor(key) {
this .key = key;
this .child = [];
}
} // Utility function to create // a new tree node function newNode(key) {
return new Node(key);
} // Function to find number of nodes // greater than x function nodesGreaterThanX(root, x) {
if (root == null )
return 0;
let count = 0;
// if current root is greater
// than x increment count
if (root.key > x)
count++;
// Number of children of root
const numChildren = root.child.length;
// recursively calling for every child
for (let i = 0; i < numChildren; i++) {
const child = root.child[i];
count += nodesGreaterThanX(child, x);
}
// return the count
return count;
} // Driver program /* Let us create below tree
* 5
* / | \
* 1 2 3
* / / \ \
* 15 4 5 6
*/
const root = newNode(5);
root.child.push(newNode(1));
root.child.push(newNode(2));
root.child.push(newNode(3));
root.child[0].child.push(newNode(15));
root.child[1].child.push(newNode(4));
root.child[1].child.push(newNode(5));
root.child[2].child.push(newNode(6));
const x = 5;
console.log( "Number of nodes greater than " + x + " are " + nodesGreaterThanX(root, x));
|
Output
Number of nodes greater than 5 are 2
Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(h) where h is the height of the binary tree.